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Surds

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Question

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${(x^{2} - \sqrt{1 - x^{2}})}^{4} + {(x^{2} + \sqrt{1 - x^{2}})}^{4}$

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Solution

We know that $a^{2} + b^{2} = {(a + b)}^{2} - 2 a b$
Take $a = {(x^{2} - \sqrt{1 - x^{2}})}^{2}$ and $b = {(x^{2} + \sqrt{1 - x^{2}})}^{2}$
Now ${(x^{2} - \sqrt{1 - x^{2}})}^{4} + {(x^{2} + \sqrt{1 - x^{2}})}^{4}$
$= ({(x^{2} - \sqrt{1 - x^{2}})}^{2} + {(x^{2} + \sqrt{1 - x^{2}})}^{2}) - 2 ({(x^{2} - \sqrt{1 - x^{2}})}^{2} {(x^{2} + \sqrt{1 - x^{2}})}^{2})$
$= x^{4} + 1 - x^{2} - 2 x \sqrt{1 - x^{2}} + x^{4} + 1 - x^{2} + 2 x \sqrt{1 - x^{2}} - 2 ({(x^{4} - 1 + x^{2})}^{2})$
$= 2 x^{4} + 2 - 2 x^{2} - 2 ({(x^{4} - 1 + x^{2})}^{2})$
$= 2 x^{4} + 2 - 2 x^{2} - 2 (x^{8} + 1 + x^{4} - 2 x^{4} - 2 x^{2} + x^{6})$
$= 2 x^{4} + 2 - 2 x^{2} - 2 x^{8} - 2 - 2 x^{4} + 4 x^{4} + 4 x^{2} - 2 x^{6}$
$= - 2 x^{8} - 2 x^{6} + 4 x^{4} + 2 x^{2}$

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